C1 June 2015 'normal' question

"Point P lies on the curve.
The normal to the curve at P is parallel to the line 2y + x = 0
Find the x coordinate of P."

I get that the gradient of the line is -1/2 but surely the gradient of the normal should be -1/2 as well if it is parallel to this line?

I really don't understand the wording of the question. The mark scheme says that the gradient you should use is 2 (which I would understand why in any other case) but the question says the normal is parallel to that line so the gradient of the normal should be the same i.e -1/2!?
If the gradient was to be 2 then you'd be using the tangent at P but how would you know in the exam to use the tangent if the question hadn't said to?

The way I intepret what you have written would lead me to think that the gradient of the normal would be -1/2 aswell.

You really need to include the full question next time, I wouldn't have known what you were asking had I not already answered this very question previously.

The derivative of a function, $\frac{dy}{dx}$ gives you the gradient of the tangent at the point $x$ on the curve $y$ by definition.

So, for example: the curve $y=x^2$ has gradient function/derivative $\frac{dy}{dx} = 2x$. That, is the gradient of the tangent at any point $x$ is $2x$. The gradient of the tangent to the curve at $x=3$ is $2(3) = 6$; this is the gradient of the tangent to the curve.

Secondly:

The normal and tangent at a point on a curve are mutually perpendicular to one another.

Hence their gradients satisfy $m_N m_T = -1$.

So, wrapping all this up:

If you are given or you found the gradient of the normal to that point, you then want to find the gradient of the tangent at that point using the perpendicularity argument. This then lets you say that $\frac{dy}{dx} = m_T = -\frac{1}{m_N}$ at the point $P$.

It is incorrect to say that $\frac{dy}{dx} = m_N$ at P, as dy/dx does not measure the gradient of the normal, it measures the gradient of the tangent. So if you want to find P, you need to use the gradient of the tangent.

Oh okay so basically to find the x coordinate of a point on a curve you always equate dy/dx to the gradient of the tangent at that point?

Oh okay so basically to find the x coordinate of a point on a curve you always equate dy/dx to the gradient of the tangent at that point?

'The curve' isn't very specific...

Sometimes tangents may involve use of the discriminant

Equal roots?